Given a (not necessarily finite dimensional) vector space $V$ prove that the center of $\operatorname{GL}(V)$ is the set of all scalar transformations (i.e all transformations of the form $a\operatorname{Id}$)?

I know how to prove this for general linear group of degree $n$, please help me solve this for the case of a general linear map.


marked as duplicate by Marc van Leeuwen linear-algebra Feb 4 '15 at 8:19

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  • $\begingroup$ I don't really understand the question: what is a "general linear group of degree $n$"? If it is $GL_n(F)$, then just notice that if $V$ is any vector space over $F$, then (assuming axiom of choice) $GL(V)$ is isomorphic to $GL_n(F)$, where $n$ is the (possibly infinite) dimension of $V$. $\endgroup$ – tomasz Oct 6 '14 at 7:38
  • $\begingroup$ @Martín-BlasPérezPinilla: Your link is misdirected, but you certainly wanted to point to the question for which this one is now marked as duplicate. Indeed the answer of Robert Israel is fine for infinite dimension, assuming (as seems necessary here) the Axiom of Choice. $\endgroup$ – Marc van Leeuwen Feb 4 '15 at 8:21
  • $\begingroup$ @MarcvanLeeuwen, thanks. The intended link was indeed math.stackexchange.com/questions/27808/…. I will delete my comment. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 4 '15 at 8:47

Let assume that there exists an element $A$ of a center of a General linear group over an arbitrary vector space $V$ such that $A$ is not a scalar transformation. (dim$V$ is infinite)

CLAIM : $\exists$ $v$ such that $v$ is not an eigenvector of $A$.

Let assume that every vector in $V$ is an eigenvector of $A$. Pick two non-parallel vectors $v_1$, $v_2$ and let $a_1$ and $a_2$ be corresponding eigenvalues. The assumption implies that $v_1+v_2$ should be an eigenvector of $A$ and let $a$ be the eigenvalue of $v_1+v_2$. $$A \cdot v_1 + A \cdot v_2 = a_1v_1 + a_2v_2 = av_1 + av_2$$ $$(a_1 - a)v_1 = (a-a_2)v_2$$ $v_1$ and $v_2$ are not parallel, so $a_1=a_2=a$ for arbitrary vector $v_1$ and $v_2$. So $A$ should be a scalar transformation. This is a contradiction. So we can assume that there exists a vector $v$ which is not an eigenvector of $A$.

Let $w = A \cdot v$, then $ w \notin <v>$. Let $S$ be a sub vector space generated by $v$, $w$. Then $\left\{ v, w \right\}$ is a basis of $S$ and there is a basis $T$ of $V$ s.t. $\left\{ v, w \right\} \subset T$. Let $S'$ be a sub vector space of $V$ which is generated by $T - \left\{ v, w \right\}$.

We can construct two general linear matrices $B_1$ and $B_2$ s.t. $$B_1 \cdot v = w, B_1 \cdot w = v$$ $$B_2 \cdot v = \frac{1}{2}w, B_2 \cdot w = v$$ $$B_1|_{S'}=B_2|_{S'}=1|_{S'}$$

Then we get following results; $$A \cdot w = A \cdot B_1 \cdot v = B_1 \cdot A \cdot v = B_1 \cdot w = v$$ $$A \cdot w = A \cdot B_2 \cdot 2v = B_2 \cdot A \cdot 2v = B_2 \cdot 2w = 2v$$ This is a contradiction. So $A$ should be a scalar transformation.

  • $\begingroup$ thank you so much, but will the maps B1, B2 be linear $\endgroup$ – Dopeman Oct 6 '14 at 10:23
  • $\begingroup$ For every vector space $V$ and it's basis $B$, we know that every $f:B \to V$ can be extended to a linear map $F: V \to V$. So we can assume that $B_1$ and $B_2$ are linear. $\endgroup$ – a-- Oct 7 '14 at 1:31

Let $T$ in the center. For any $L$ we have $T\circ L= L\circ T$, that is $$T(Lx) = L(Tx)$$ for all $L$ and all $x \in V$.

Let $x$ in $V$. There exists $L$ linear map so that the subspace $\{y \ | \ Ly = y\}$ equals $\mathbb{F} \cdot x$ ( use a basis starting from $x$).

We get $L(Tx) = T(Lx) = Tx$ and so $Tx \in \mathbb{F} \cdot x$.

So, for any $x\in V$ we have $T(x)$ proportional to $x$. It's easy now.


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